Given a string, determine if it is a palindrome, considering only alphanumeric characters and ignoring cases.
For example,
“A man, a plan, a canal: Panama” is a palindrome.
“race a car” is not a palindrome.
Note:
Have you consider that the string might be empty? This is a good question to ask during an interview.
For the purpose of this problem, we define empty string as valid palindrome.
翻译：
给定一个字符串，检测它是否是回文对称的，只考虑其中字母与数字的字符。
例如”A man, a plan, a canal: Panama”是回文对称的，”race a car”不是。
提示：你考虑过字符串可能是空的情况吗？这是一个面试中应该问出的好问题。为了处理这个问题，我们假定空串是回文对称的。
判断回文数（字符串里有字符、数字、标点符号，需要去掉标点符号）
import java.util.*;
public class PalindromeNumber {
public static void main(String[] args) {
// TODO Auto-generated method stub
//Scanner scan = new Scanner(System.in);
//System.out.println("请输入数字:");
//String strNum = scan.next();
String strNum="q1Qq1q";
boolean result = isPalindrome(strNum);
System.out.println(result);
}
public static Boolean isPalindrome(String str){
char[] charArr=str.toCharArray();
char[] ch=new char[charArr.length];
int fix='a'-'A';
for(int i=0;i<=charArr.length-1;i++){
if((charArr[i]>='a'&&charArr[i]<='z')||(charArr[i]>='0'&&charArr[i]<='9')){
ch[i]=charArr[i];
}else if ((charArr[i]>='A'&&charArr[i]<='Z'))
ch[i]=(char)(charArr[i]+fix);
}
boolean result=false;
int m=0;
int n=ch.length-1;
while(m<=n){
if(ch[m]==ch[n]){
result=true;
m++;
n--;
}else {
return result=false;
}
}
return result;
}
}